Problem: ${\sqrt[3]{729} = \text{?}}$
Explanation: $\sqrt[3]{729}$ is the number that, when multiplied by itself three times, equals $729$ If you can't think of that number, you can break down $729$ into its prime factorization and look for equal groups of numbers. So the prime factorization of $729$ is $3\times 3\times 3\times 3\times 3\times 3$ We're looking for $\sqrt[3]{729}$ , so we want to split the prime factors into three identical groups. Notice that we can rearrange the factors like so: $729 = 3\times 3\times 3\times 3\times 3\times 3 = \left(3\times 3\right)\times\left(3\times 3\right)\times\left(3\times 3\right)$ So $\left(3\times 3\right)^3 = 9^3 = 729$ So $\sqrt[3]{729}$ is $9$.